# Need urgent help with proof involving summation with “n choose k” please! [duplicate]

Possible Duplicate:
Crazy induction

I'm in urgent need of help solving this proof. I have my final exam tomorrow, and me passing depends on me being able to understand how to prove this statement, and similar statements. Can anyone explain to me how this is proved?

$$\displaystyle \sum_{i = 0}^n {n \choose i}{i^2}= (n+1)(n){2^{n-2}}$$ for n$\ge0$

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## marked as duplicate by Marvis, amWhy, Mike Spivey, draks ..., Davide GiraudoDec 10 '12 at 20:02

This question was marked as an exact duplicate of an existing question.

That title @Marvis – Jean-Sébastien Dec 10 '12 at 19:42

We know the nbinomial expansion $$f(x):=\sum_{i=0}^n {n\choose i}x^i = (1+x)^n.$$ Note that taking the derivative and multiplying by $x$ turns $x^i$ into $ix^i$, thus $$xf'(x)=\sum_{i=0}^n {n\choose i}ix^i$$ and repeating this step $$x (f'(x)+xf''(x))=\sum_{i=0}^n {n\choose i}i^2x^i$$ We want this evaluated at at $x=1$. With $f'(x) = n(1+x)^{n-1}$ and $f''(x)=n(n-1)(1+x)^{n-2}$, we obtain $$1\cdot(n2^{n-1}+1\cdot n(n-1)2^{n-2})=n(n-1+2)2^{n-2}=n(n+1)2^{n-2}.$$